STEORN DEMO LIVE & STREAM in Dublin, December 15th, 10 AM

[callanan]

Sorry Ossie and Ben, one more thing. What's the diameter of your rotor?

Mine is 100mm or 4 inches. I used two disks used for cutting metal with angle grinders. Standard 4 inch angle grinder disks.

Ossie

[k4zep]

Thanks..
Surely, if the shunt is exactly 1 ohm, the scope will read the current directly. Here it is 0.01. I'm a bit confused, because adjusting the scope reading gives 1800 Amp as peak current which is absurd.

Be sure that resistor is .01 OHM.  You can verify that resistance using ohms law and a constant current power supplly.  Put 1 amp into resistor and measure voltage across it.  IF 10 mv, .01 ohm, if 100 mv, .1 ohm, if 1 ohm, will be 1 VOLT.  E=IR.  We have a known current and a unknown resitance......Measured E is 10 mv, therefore R=E/I.........OK  NOW that you are sure of your resistance.......

Ohms law. E (across shunt)=I (current thrugh shunt) X (times) R (resistance of shunt) or E=IR.  IF current is 1 amp and resistance is .01 ohm, voltage measured across resistor will be .01 Volts. (10 mv.)  Now you know the voltage across shunt per. amp.  Set your scope  Vert. to 10 mv/div and each amp will be one division or 100 mv/div and then 10 amps will be one division......  The important thing is to know how many mv is developed across that shunt/ amp.  In this case 10 mv! SCOPE always ground to ground side of power supply.   OK..... Instrumentation 101

Ben

[k4zep]

Sorry Ossie and Ben, one more thing. What's the diameter of your rotor?

Sprite drink al can.......aaaaaaaaaaaaaaaaaaaa   hummmmmmmm................How bout 65 mm. give or take 68 with magnets.

[Omnibus]

Sorry Ossie, but yours appears to be undeunity (if the unaccounted for heat losses don't turna that around). So, let's review what you have:

Input energy every second:
E = 2.2V
I = 0.3A
duty cycle = 0.48
Therefore, energy spent every secon is:
W = 0.48 x 2.2V x 0.3A = 0.32W or 0.32J every second

Output energy (only rotational kinetic energy):
Mass of rotor = 0.35kg
Radius of rotor = 0.05m
Rotations per second = 180/60 = 3rps
KE = 0.5 x 0.35kg x 0.05m x 0.05m x (2 x 3.14 x 3)^2 = 0.16J

Therefore, efficiency = 0.16/0.32 = 0.5 (heat losses unaccounted for) which is far from OU.

[Omnibus]

Hi Ben,

This is what appears to come out from your data:

Input energy every second:
E = 10V
I = 0.4A
duty cycle = 0.5
Therefore, energy spent every secon is:
W = 0.5 x 10V x 0.4A = 2W or 2J every second

Output energy (only rotational kinetic energy):
Mass of rotor = 0.033kg
Radius of rotor = 0.034m
Rotations per second = 800/60 = 13.3rps
KE = 0.5 x 0.033kg x 0.034m x 0.034m x (2 x 3.14 x 13.3)^2 = 0.13J

Therefore, efficiency = 0.13/2 = 0.065 (heat losses unaccounted for) which is even farther from OU than Aussie's.

There's an almost an order of magnitude discrepancy between yours and Aussies' so maybe some of the data aren't correct. Could you please double check just to make sure. Thanks.